Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 175 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places.

Answer 1

The 99% confidence interval is
`0.3003 < I < 0.3997`

Explanation:

From the question we are told that

The sample size is
`n = 500`

The the number that are parents x = 175

The proportion of parents is mathematically represented as

`\r p = (x)/(n)`

substituting values

`\r p = (175)/(500)`

`\r p = 0.35`

The level of confidence is given as 99% which implies that the level of significance is

`\alpha = 100 - 99`

`\alpha =`1%

`\alpha = 0.01`

The critical value for this level of significance is obtained from the table of critical value as

`t_(x, \alpha ) = t_(175, 0.05) = 2.33`

Generally the margin of error is mathematically evaluated as

`M =( t_(175, 0.01 ) * √(\r p (1-\r p)) )/(√(n) )`

substituting values

`M =( 2.33 * √(\r 0.35 (1-0.35)) )/(√(500) )`

`M = 0.0497`

Generally the 99% confidence interval is mathematically represented as

`I = \r p \pm M`

`\r p -M < I < \r p + M`

substituting values

`0.35 -0.0497 < I < 0.35 + 0.0497`

`0.3003 < I < 0.3997`